Sum of the rows of Pascal's Triangle.

Question

I've discovered that the sum of each row in Pascal's triangle is $2^n$, where $n$ number of rows. I'm interested why this is so. Rewriting the triangle in terms of C would give us $0C0$ in first row. $1C0$ and $1C1$ in the second, and so on and so forth. However, I still cannot grasp why summing, say, 4C0+4C1+4C2+4c3+4C4=2^4.

Answer 1

There are various different ways to look at this. Here's one:

Two adjacent numbers in a row get added to get the number in the row below it: $$ \begin{array}{cccccccccc} & & 1 & & & & & 8 & & & & 28 & & & & 56 & & & & 70 & & \cdots \\ & & & & & & & & \searrow & & \swarrow \\ 1 & & & & & 9 & & & & 36 & & & & 84 & & & & 126 & & \cdots & & & & \cdots \end{array} $$ That means every number in a row is added into the next row twice: $$ \begin{array}{cccccccccc} & & 1 & & & & & 8 & & & & 28 & & & & 56 & & & & 70 & & \cdots \\ & & & & & & \swarrow & & \searrow \\ 1 & & & & & 9 & & & & 36 & & & & 84 & & & & 126 & & \cdots & & & & \cdots \end{array} $$ Since every number is added into the next row twice, the sum of the numbers in the next row is twice as big.

Here's another: In row $9$ (which is the tenth row, since the first row is "row $0$), the entries are. $$ \binom 9 0 = 1,\ \binom 9 1 = 9,\ \binom 9 2 = 36,\ \binom 9 3 = 84,\ \binom 9 4 = 126,\ \ldots $$ These are

• the number of subsets of size $0$ of a set of size $9$, and
• the number of subsets of size $1$ of a set of size $9$, and
• the number of subsets of size $2$ of a set of size $9$, and
• the number of subsets of size $3$ of a set of size $9$, and
• and so on.

Their sum is therefore the total number of subsets of a set of size $9$. If you know that that is $2^9$, you've got it.

Answer 2

If you know the binomial theorem, then it's easy: consider $(1+1)^n$.

If you understand $nCk$ in combinatorial terms as the number of subsets of $k$ elements chosen from a universe of $n$ elements, then it's easy because the total number of subsets is $2^n$.

Answer 3

Use the binomial theorem as:

$$(1+x)^n = {n\choose 0} + {n\choose 1}x^1+ {n\choose 2}x^2+\cdots+{n\choose n}x^n$$ Put $x=1$ to get $$2^n= {n\choose 0} + {n\choose 1}+ {n\choose 2}+\cdots+{n\choose n}.$$

Answer 4

This can also be proven by induction. In order to do so, it is very helpful to prove the following identity: $$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$$ This can be proven combinatorially, but for the sake of expediency, \begin{align*} \binom{n}{k} + \binom{n}{k+1} &= \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!} \\ &= \frac{n!}{k!(n-k)(n-k-1)!} + \frac{n!}{(k+1)k!(n-k-1)!} \\ &= \frac{n!}{k!(n-k-1)!}\left(\frac{1}{n-k} + \frac{1}{k+1}\right) \\ &= \frac{n!}{k!(n-k-1)!}\left(\frac{k+1}{(n-k)(k+1)} + \frac{n - k}{(n-k)(k+1)}\right) \\ &= \frac{n!}{k!(n-k-1)!} \cdot \frac{n+1}{(n-k)(k+1)} \\ &= \frac{(n+1)n!}{(k+1)k!(n-k)(n-k-1)!} \\ &= \frac{(n+1)!}{(k+1)!(n-k)!} \\ &= \binom{n+1}{k+1}. \end{align*}

Now, let's launch into the induction proof. First, note that $\binom{0}{0} = 1 = 2^0$, so the base case holds.

Next, assume that $\binom{n}{0} + \ldots + \binom{n}{n} = 2^n$. Then, using the identity, \begin{align*} &\binom{n+1}{0} + \binom{n+1}{1} + \binom{n+1}{2} + \ldots + \binom{n+1}{n} + \binom{n+1}{n+1} \\ = \, &\binom{n+1}{0} + \left(\binom{n}{0} + \binom{n}{1} \right) + \left(\binom{n}{1} + \binom{n}{2} \right) + \ldots + \left(\binom{n}{n-1} + \binom{n}{n} \right) + \binom{n+1}{n+1}. \end{align*} Note that each term of the form $\binom{n}{k}$ where $1 \le k \le n-1$ occurs exactly twice in the above sum. Also, the other terms, $\binom{n+1}{0}, \binom{n}{0}, \binom{n+1}{n+1}, \binom{n}{n}$ are all $1$. Thus we have, \begin{align*} &\binom{n+1}{0} + \binom{n+1}{1} + \binom{n+1}{2} + \ldots + \binom{n+1}{n} + \binom{n+1}{n+1} \\ = \, &2\left(\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n-1} + \binom{n}{n} \right) = 2 \cdot 2^n = 2^{n+1}. \end{align*}

Answer 5

Here's a simple combinatorial proof that is mentioned above but not actually spelled out, whose full exposition — if you are asking this question in the first place — may not be obvious. (But the other detailed answers are much more complicated than is necessary, I think.)

Suppose you flip a coin $n$ times. You want to work out the number of ways that that can occur. The most intuitive answer is $2^n$, a permutation where we have $2$ outcomes possible, we're allowed to repeat, and we have a sequence of length $n$.

But, an alternative method would be to realize that the number of ways to choose $0, 1, 2, ... n$ trials to be heads must cover all possible outcomes, and thus

\begin{align*} 2^n = \sum_{j = 0}^n \binom{n}{j}. \end{align*}

If you know the connection between Pascal's triangle and the binomial coefficient, the proof is complete. If you don't, then simply consider this bonus proof.

1. Each term in the triangle is the sum of the two above it.
2. $\binom{n-1}{k-1} + \binom{n-1}{k} = \binom{n}{k}$ since the number of ways to pick a subset $k$ from a group of $n$ must be the number of ways to pick a group that definitely involves one element $j$, which is $\binom{n-1}{k-1}.$ (We have one fewer option and need to pick a group that is one smaller.) Add to this the number of ways to pick a group that definitely does not involve that element, $\binom{n-1}{k}.$ (We have one fewer option but need the same number of elements.)
3. Thus, if we interpret $n$ as the row number and $k$ as the number along the row, we see that the binomial coefficient is also the sum of the two coefficients "above" it.
4. There are two caveats. First, the actual row number is $n+1$ since the first row should be interpreted as $\binom{0}{0}$ since we want to make the two triangles agree (note that if we tried to, for example, make $\binom{1}{0}$ or $\binom{1}{1}$ the first row, we couldn't pick which since both $=1$). The second caveat, relatedly, is that in both triangles, the first row "comes from nowhere" and we posit zeros outside the triangle. In the combinatorial triangle, we can interpret the first row as also coming from nowhere and assign entries outside the triangle, generically of the form $\binom{n}{-1}$ and $\binom{n}{n+1}$ as being zero on logical grounds.

So, the sum of each row in the triangle is also the sum of the binomial coefficients given above.